Consider the Lagrangian for a simple harmonic oscillator, written in terms of its "natural" variables, position $x$ and velocity $\textrm{v}$:
$$
L(x, \textrm{v}) = \frac{1}{2}m\textrm{v}^2 - \frac{1}{2}kx^2\;.
$$
We can construct another quantity, also a function of $x$ and $\textrm{v}$, which I will call the "velocity form of the energy" $W$:
$$
W(x, \textrm{v}) = \frac{\partial L}{\partial \textrm{v}}\textrm{v} - L \tag{A}
$$
$$= \frac{1}{2}m\textrm{v}^2 + \frac{1}{2}kx^2\;.
$$
Note that $W$ is of interest since $W(x(t),\dot x(t))$ is constant in time when $x(t)$ is the true classical path.
This thing I'm calling the "velocity form of the energy" $W$ is also called the "Hamiltonian" $H$, when written as a function of the position $x$ and momentum $p$.
The definition of the momentum, $p$, is:
$$
p \equiv \frac{\partial L}{\partial \textrm{v}} = \frac{\partial W}{\partial \textrm{v}} = m\textrm{v}\;.
$$
The second equality follows from the fact that $x$ and $\textrm{v}$ are independent variables and the further requirement that the kinetic energy is a quadradic function of the velocity. (See section below for more details about this.)
In more detail, given a kinetic energy, $T(\textrm{v})$, and a potential energy, $U(x)$, we have:
$$
p=\frac{\partial L}{\partial \textrm{v}}=\frac{\partial }{\partial \textrm{v}}\left(T(\textrm{v}) - U(\textrm{x})\right)
$$
$$
=\frac{\partial T}{\partial \textrm{v}}\;.
$$
We also have, by the definition in Eq. (A) above of the velocity form of the energy,
$$
\frac{\partial W}{\partial \textrm{v}} = \textrm{v}\frac{\partial^2 L}{\partial \textrm{v}^2} = \textrm{v}\frac{\partial^2 T}{\partial \textrm{v}^2}\;.
$$
If $T(\textrm{v})$ is a homogeneous power of $\textrm{v}$, we can write
$$
T = av^b\;,
$$
where $a$ and $b$ are constants and $a$ is not zero. We thus have
$$
\frac{\partial T}{\partial \textrm{v}} = abv^{(b-1)}\tag{1}
$$
and
$$
\textrm{v}\frac{\partial^2 T}{\partial \textrm{v}^2} = ab(b-1)v^{(b-1)}\tag{2}
$$
and we have equality of the right-hand sides of Eqs. (1) and (2) when either $b=0$ (which is not interesting) or when
$$
b=2\;,
$$
which is typical for a kinetic energy term.
Given our expression for $p(\textrm{v})=m\textrm{v}$, we can invert the expression to find $\textrm{v}(p) = p/m$, and we can write the Hamiltonian energy as:
$$
H(x, p) = W(x, \textrm{v}(p)) = \frac{p^2}{2m}+\frac{1}{2}kx^2\;.
$$
This is "all" that Fermi has done, since Fermi's system is "just" a collection of uncoupled simple harmonic oscillators. So, he is just working in complete analogy to the simple harmonic oscillator.
In other words, for each mode, labeled by $s$, in the expression
$$
W_e = \frac{\Omega}{8\pi c^2}\sum_s
\left(\frac{1}{2}\dot u_s^2 + 2\pi^2\nu_s^2 u_s^2\right)
$$
we can make the following identifications, in analogy to the first section of this answer:
$$
\dot u_s \to \textrm{v}_s
$$
$$
u_s \to x_s
$$
$$
\frac{\Omega}{8\pi c^2} \to m_s
$$
and
$$
\frac{\Omega}{8\pi c^2}4\pi^2\nu_s^2 \to k_s\;,
$$
in which case we have, by the definitions above, the velocity form of the energy:
$$
W_e(x_1,\ldots,\textrm{v}_N) = \sum_s \frac{1}{2}m_s \textrm{v}_s^2 + \frac{1}{2}k_s x_s^2
$$
And, also in analogy to the first section of this answer, we can make the identifications:
$$
p = \frac{\partial W}{\partial \textrm{v}}\to p_s = \frac{\partial W_e}{\partial \dot u_s}=\frac{\Omega}{8\pi c^2}\dot u_s = m_s \textrm{v}_s
$$
Now, by substituting into the velocity form $W_e(x_1,\ldots,\textrm{v}_N)$ with $v_s(p_s)$ for $v_s$ we arrive at the Hamiltonian form of the energy (Fermi uses the same symbol $W_e$ for it):
$$
W_e(x_1,\ldots,p_N) =
\sum_s \frac{p_s^2}{2m_s}+\frac{1}{2}k_s x_s^2
=\sum_s \underbrace{\frac{8\pi^2 c^2}{\Omega}}_{1/m_s}\frac{p_s^2}{2} + \frac{1}{2}\underbrace{\frac{\Omega}{8\pi c^2}4\pi^2\nu_s^2}_{k_s} u_s^2\;,
$$
which is what Fermi arrived at. (Note that Fermi used a more-curly "$v$" symbol for momentum instead of the "$p$" symbol I am using. I find the more curly "$v$" to be a little confusing here, since such a symbol is often used for velocity (rather than momentum) and also it looks fairly similar to Fermi's "nu" symbol $\nu$. So, I am using the symbol $p_s$ for the momentum of a mode, and I am using a non-curly $\textrm{v}_s$ for velocity.)
